Latest TCS Aptitude Question SOLUTION: In the reading room of a library, there are23 reading spots. Each reading spot consists of a round table with 9 chairs placed around it. There are some readers such that in each o

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In the reading room of a library, there are23 reading spots. Each reading spot consists of a round table with 9 chairs placed around it. There are some readers such that in each occupied reading spot there are different numbers of readers. If in all there are 36 readers, how many reading spots do not have even a single reader?
a) 8 b) None c) 16 d) 15

as there are 23 spots so,the there are 23 tables and each table there are 9 chairs.the condition is:each table occupies different no. of people.and as there are 36 people so these people can be arranged in each table:1,2,3,4,5,6,7,8 in this way.that means 36=1+2+3+4+5+6+7+8.that means 8 tables are occupied.so.(23-8)=15 tables have no single readers..

there are 23 reading spot and max limit to people sit on 1 spot is 9.
then answer may be-
1. 1+2+3+4+5+6+7+8=36 people
remaining empty table=23-8=15 ans
2.9+8+7+6+5+1=36 people can sit
then rem. empty table=23-6=17 ans

Hey my sincere advise to m4maths..please try to post the actual answer explanation when ur going to upload any questions.then only we are able to get clarification on any problem..because every one has their own views...so automatically it leads to the wrong way...

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In the reading room of a library, they are 23 reading spots. Each reading spot consists of a round table with 9 chairs placed around it. There some readers such that in each occupaid reading spot there are diffirent number of readers. If in all there are 36 readers, how many reading spots do not have even a single reader?

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In a reading room of a library there are 22 reading spots. Each reading spot consists of a round table with 9 chairs placed around it. In each table different numbers of people are seated. If in all there are 36 readers how many reading spots will be left out without at least 1 reader?

a) 8
b) 12
c) 16
d) None of these

Since at least there can be 1 reader in a table so the following case is possible
1st table- 9 members
2nd table- 8 members
3rd table- 7 members
4th table- 6 members
5th table - 5 members
6th table -1 member
therefore 9+8+7+6+5+1=36 and no. of tables are 6. so, no. of remaining tables are 22-6 = 16(ans).

Since at least there can be 1 reader in a table so the following case is possible
1st table- 8 members
2nd table- 7 members
3rd table- 6 members
4th table- 5 members
5th table - 4 members
6th table - 3 members
7th table - 2 members
8th table - 1 member
therefore 8+7+6+5+1=36 and no. of tables are 5. so, no. of remaining tables are 22-5 = 17(ans).

since different number of people are seated since atleast1 reader should not be common 1 reader should not present as common in each reading spot---=arrange 9 difeerent people in 4 reading spots =9*4=36 and remaining are 22-4=18 so d is answer